# Monday April 27th

We get a non-split SES
\begin{align*}
0 \to M(xs\cdot \lambda) \to T_\mu^\lambda M(x\cdot \mu) \to M(x\cdot \lambda) \to 0
.\end{align*}

Since $w\cdot \mu$ is the highest weight of $T(w\cdot \mu)$ and occurs with multiplicity one, apply this to all Verma section $M(x\cdot \mu)$, including $x= w$ in a standard filtration of $T(w\cdot \mu)$, thus $T_\mu^\lambda T(w\cdot \mu)$ has highest weight $ws\cdot \lambda$ with multiplicity one.

By Prop 11.1e and Theorem 11.2, $T_\mu^\lambda T(w\cdot \mu) \cong T(ws \cdot \lambda) \oplus T$ where $T$ is a tilting module in $\OO_{\chi_\lambda}$ having all weights less than $ws\cdot \lambda$. 
It suffices to show $T=0$, or equivalently $T_\lambda^\mu T = 0$.

Lemma (Translating and Inverting Doubles the Character)
: For $\lambda, \mu$ as above and any $M\in \OO_{\chi_\mu}$, $\ch T_\lambda^\mu T_\mu^\lambda M = 2 \ch M$.

Proof:
By writing $\ch M$ in a basis of $M(x\cdot \mu)$ with $x\in W$ and $xs > x$, since $M(xs \cdot \mu) = M(x \cdot \mu)$, it suffices to show this for $M = M(x\cdot \mu)$.
But $T_\lambda^\mu T_\mu^\lambda M(x\cdot \mu)$ is given by applying $T_\lambda^\mu$ to the original SES and we know
$$
T_\lambda^\mu M(xs\cdot \lambda) = M(xs \cdot \mu) = M(x\cdot \mu) = T_\lambda^\mu M(x\cdot \lambda)
$$
Thus $\ch T_\lambda^\mu T_\mu^\lambda M(x\cdot \mu) = 2\ch M(x\cdot \mu)$.

$\qed$

Now by the lemma, $T_\lambda^\mu T(ws \cdot \lambda) \oplus T_\lambda^\mu T = T_\lambda^\mu T_\mu^\lambda T(w\cdot \mu)$ has formal character $2 \ch T(w\cdot \mu)$.
Since it's a tilting module,we must have $T_\lambda^\mu T_\mu^\lambda T(w\cdot \mu) = T(w\cdot \mu) \oplus T(w\cdot \mu)$.
In particular, it has highest weight $w\cdot \mu$ with multiplicity 2.

If we can show that $T_\lambda^\mu T(ws\cdot \lambda)$ already has $w\cdot \mu$ as a weight with multiplicity 2, it will follow that the remaining term must be zero as desired.

Start with an embedding $\phi: M(ws\cdot \lambda) \injects T(ws\cdot \lambda)$.
Using Theorem 6.13c, our $\ext^1$ vanishes between Vermas and dual Vermas, and so we have $\ext^1 (T(ws\cdot \lambda), M(w\cdot\lambda A)\dual) = 0$.

Dualizing, $\ext^1(M(w\cdot \lambda), T(ws\cdot \lambda)) = 0$ and this sequence must split.
Applying $\hom(\wait, T(ws\cdot \lambda))$, we get a LES
\begin{align*}
\hom(T_\mu^\lambda M(w\cdot \lambda), T(ws\cdot \lambda)) \to \hom(M(ws\cdot \lambda), T(ws\cdot \lambda)) \to \ext^1(M(w\cdot \lambda), T(ws\cdot\lambda)) \to \cdots
.\end{align*}

Since the last term vanishes, a $\phi$ in the middle term lifts to the first term.

Proposition (Injective Embedding of Vermas into Tilting Modules)
: $(\ker\phi)_{w\cdot \lambda} = 0$.

Proof
:   If not, since $\phi$ restricted to $M(ws\cdot \lambda)$ is injective, and using the origin SES we must have a preimage $v\in (\ker \phi)_{w\cdot \lambda}$ of the highest weight vector in $M(w\cdot \lambda)$.

    But every $z\in T_\mu^\lambda  M(w\cdot \mu)$ can be written uniquely (since the SES splits as vector spaces) as $z = uv + m$ where $u\in U(\lien^-)$ and $m\in M(ws\cdot \lambda)$.
    Then $\phi(z) = 0 + \phi(m) = m \in M(ws\cdot \lambda) \subset T(ws\cdot\lambda)$, since $\phi$ is the identity on this submodule.
    But then $\phi$ provides a splitting of the non-split SES, a contradiction.

Thus $\phi$ induces a nonzero homomorphism
\begin{align*}
\bar \phi : M(w\cdot \lambda) \cong T_\mu^\lambda M(w\cdot \mu) / M(ws\cdot \lambda) \to T(ws\cdot \lambda) / M(ws\cdot \lambda)
.\end{align*}

In particular, $w\cdot \lambda$ is a weight of the quotient module, and is a maximal weight -- keeping in mind that $T(ws\cdot \lambda)$ has a standard filtration with sections $M(x\cdot \lambda)$ for $x \leq ws$ with $M(ws\cdot \lambda)$ occurring just once.
The quotient module also has a standard filtration, so $M(w\cdot \lambda)$ must occur in the standard filtration of $T(ws\cdot \lambda)$ along with $M(ws\cdot \lambda)$.

Since $w<ws$, theorem 1.4 provides an injection $M(w\cdot \lambda) \injects M(ws\cdot \lambda)$.
Applying $T_\lambda^\mu$ to these two copies of $M(w\cdot \lambda)$, we get $w\cdot \mu$ a weight with multiplicity at least 2.

$\qed$

Corollary (Standard Multiplicities of Vermas in Tilting Modules)
: Under the hypotheses of the theorem, $(T(ws\cdot \lambda): M(xs \cdot \lambda)) = (T(ws\cdot \lambda) : M(x\cdot \lambda))$.

Proof
:   $T(ws\cdot \lambda) = T_\mu^\lambda T(w\cdot \mu)$, and wlog $x < xs$ since the claimed formula is symmetric in $x$ and $xs$ and $xs\cdot \mu = x\cdot \mu$.
    Now tilting modules (and thus their filtration multiplicities) are determined by their formal characters.
    Using 1', we see that each occurrence of $M(x\cdot \mu)$ as a section of $T(w\cdot \mu)$ leads to exactly one occurence of $M(xs\cdot \lambda)$ and $M(x\cdot \lambda)$ in the character of $T(ws \cdot \lambda)$.

Remark:
Both the theorem and corollary can fail when $ws<w$.